# 1 1 Slide MA4704Gerry Golding Developing Null and Alternative Hypotheses Hypothesis testing can be used to determine whether Hypothesis testing can be.

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1 1 Slide MA4704Gerry Golding Developing Null and Alternative Hypotheses Hypothesis testing can be used to determine whether Hypothesis testing can be used to determine whether a statement about the value of a population parameter a statement about the value of a population parameter should or should not be rejected. should or should not be rejected. The null hypothesis, denoted by H 0, is a tentative The null hypothesis, denoted by H 0, is a tentative assumption about a population parameter. assumption about a population parameter. The alternative hypothesis, denoted by H a, is the The alternative hypothesis, denoted by H a, is the opposite of what is stated in the null hypothesis. opposite of what is stated in the null hypothesis. The alternative hypothesis is what the test is The alternative hypothesis is what the test is attempting to establish. attempting to establish.

2 2 Slide MA4704Gerry Golding n Testing Research Hypotheses Developing Null and Alternative Hypotheses The research hypothesis should be expressed as The research hypothesis should be expressed as the alternative hypothesis. the alternative hypothesis. The conclusion that the research hypothesis is true The conclusion that the research hypothesis is true comes from sample data that contradict the null comes from sample data that contradict the null hypothesis. hypothesis.

3 3 Slide MA4704Gerry Golding Developing Null and Alternative Hypotheses n Testing the Validity of a Claim Manufacturers’ claims are usually given the benefit Manufacturers’ claims are usually given the benefit of the doubt and stated as the null hypothesis. of the doubt and stated as the null hypothesis. The conclusion that the claim is false comes from The conclusion that the claim is false comes from sample data that contradict the null hypothesis. sample data that contradict the null hypothesis.

4 4 Slide MA4704Gerry Golding n Testing in Decision-Making Situations Developing Null and Alternative Hypotheses A decision maker might have to choose between A decision maker might have to choose between two courses of action, one associated with the null two courses of action, one associated with the null hypothesis and another associated with the hypothesis and another associated with the alternative hypothesis. alternative hypothesis. Example: Accepting a shipment of goods from a Example: Accepting a shipment of goods from a supplier or returning the shipment of goods to the supplier or returning the shipment of goods to the supplier supplier

5 5 Slide MA4704Gerry Golding One-tailed(lower-tail)One-tailed(upper-tail)Two-tailed Summary of Forms for Null and Alternative Hypotheses about a Population Mean n The equality part of the hypotheses always appears in the null hypothesis. in the null hypothesis. In general, a hypothesis test about the value of a In general, a hypothesis test about the value of a population mean  must take one of the following population mean  must take one of the following three forms (where  0 is the hypothesized value of three forms (where  0 is the hypothesized value of the population mean). the population mean).

6 6 Slide MA4704Gerry Golding The director of medical services The director of medical services wants to formulate a hypothesis test that could use a sample of emergency response times to determine whether or not the service goal of 12 minutes or less is being achieved. n Example: Metro EMS Null and Alternative Hypotheses

7 7 Slide MA4704Gerry Golding Null and Alternative Hypotheses The emergency service is meeting the response goal; no follow-up action is necessary. The emergency service is not meeting the response goal; appropriate follow-up action is necessary. H 0 :  H a :  where:  = mean response time for the population of medical emergency requests of medical emergency requests

8 8 Slide MA4704Gerry Golding Type I Error Because hypothesis tests are based on sample data, Because hypothesis tests are based on sample data, we must allow for the possibility of errors. we must allow for the possibility of errors. n A Type I error is rejecting H 0 when it is true. n The probability of making a Type I error when the null hypothesis is true as an equality is called the null hypothesis is true as an equality is called the level of significance. level of significance. n Applications of hypothesis testing that only control the Type I error are often called significance tests. the Type I error are often called significance tests.

9 9 Slide MA4704Gerry Golding n Example: Metro EMS Null and Alternative Hypotheses Operating in a multiple Operating in a multiple hospital system with approximately 20 mobile medical units, the service goal is to respond to medical emergencies with a mean time of 12 minutes or less. A major west coast city provides A major west coast city provides one of the most comprehensive emergency medical services in the world.

10 Slide MA4704Gerry Golding Type II Error n A Type II error is accepting H 0 when it is false. n It is difficult to control for the probability of making a Type II error. a Type II error. n Statisticians avoid the risk of making a Type II error by using “do not reject H 0 ” and not “accept H 0 ”. error by using “do not reject H 0 ” and not “accept H 0 ”.

11 Slide MA4704Gerry Golding Type I and Type II Errors CorrectDecision Type II Error CorrectDecision Type I Error Reject H 0 (Conclude  > 12) Accept H 0 (Conclude  < 12) H 0 True (  < 12) H 0 False (  > 12) Conclusion Population Condition

12 Slide MA4704Gerry Golding p -Value Approach to One-Tailed Hypothesis Testing Reject H 0 if the p -value < . Reject H 0 if the p -value < . The p -value is the probability, computed using the The p -value is the probability, computed using the test statistic, that measures the support (or lack of test statistic, that measures the support (or lack of support) provided by the sample for the null support) provided by the sample for the null hypothesis. hypothesis. If the p -value is less than or equal to the level of If the p -value is less than or equal to the level of significance , the value of the test statistic is in the significance , the value of the test statistic is in the rejection region. rejection region.

13 Slide MA4704Gerry Golding n p -Value Approach p -value  p -value  0 0 - z  = -1.28 - z  = -1.28  =.10 z z z = -1.46 z = -1.46 Lower-Tailed Test About a Population Mean:  Known Sampling distribution of Sampling distribution of p -Value < , so reject H 0. Signal: Noise

14 Slide MA4704Gerry Golding n p -Value Approach p -Value  p -Value  0 0 z  = 1.75 z  = 1.75  =.04 z z z = 2.29 z = 2.29 Upper-Tailed Test About a Population Mean:  Known Sampling distribution of Sampling distribution of p -Value < , so reject H 0.

15 Slide MA4704Gerry Golding Critical Value Approach to One-Tailed Hypothesis Testing The test statistic z has a standard normal probability The test statistic z has a standard normal probability distribution. distribution. We can use the standard normal probability We can use the standard normal probability distribution table to find the z -value with an area distribution table to find the z -value with an area of  in the lower (or upper) tail of the distribution. of  in the lower (or upper) tail of the distribution. The value of the test statistic that established the The value of the test statistic that established the boundary of the rejection region is called the boundary of the rejection region is called the critical value for the test. critical value for the test. n The rejection rule is: Lower tail: Reject H 0 if z < - z  Lower tail: Reject H 0 if z < - z  Upper tail: Reject H 0 if z > z  Upper tail: Reject H 0 if z > z 

16 Slide MA4704Gerry Golding  0 0  z  =  1.28 Reject H 0 Do Not Reject H 0 z Sampling distribution of Sampling distribution of Lower-Tailed Test About a Population Mean:  Known n Critical Value Approach

17 Slide MA4704Gerry Golding  0 0 z  = 1.645 Reject H 0 Do Not Reject H 0 z Sampling distribution of Sampling distribution of Upper-Tailed Test About a Population Mean:  Known n Critical Value Approach

18 Slide MA4704Gerry Golding Steps of Hypothesis Testing Step 1. Develop the null and alternative hypotheses. Step 2. Specify the level of significance . Step 3. Collect the sample data and compute the test statistic. p -Value Approach Step 4. Use the value of the test statistic to compute the p -value. p -value. Step 5. Reject H 0 if p -value < .

19 Slide MA4704Gerry Golding Critical Value Approach Step 4. Use the level of significance  to determine the critical value and the rejection rule. Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H 0. rule to determine whether to reject H 0. Steps of Hypothesis Testing

20 Slide MA4704Gerry Golding n Example: Metro EMS The EMS director wants to The EMS director wants to perform a hypothesis test, with a.05 level of significance, to determine whether the service goal of 12 minutes or less is being achieved. The response times for a random The response times for a random sample of 40 medical emergencies were tabulated. The sample mean is 13.25 minutes. The population standard deviation is believed to be 3.2 minutes. One-Tailed Tests About a Population Mean:  Known

21 Slide MA4704Gerry Golding 1. Develop the hypotheses. 2. Specify the level of significance.  =.05 H 0 :  H a :  p -Value and Critical Value Approaches p -Value and Critical Value Approaches One-Tailed Tests About a Population Mean:  Known 3. Compute the value of the test statistic. Signal: Noise

22 Slide MA4704Gerry Golding 5. Determine whether to reject H 0. We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes. p –Value Approach p –Value Approach One-Tailed Tests About a Population Mean:  Known 4. Compute the p –value. For z = 2.47, cumulative probability =.9932. p –value = 1 .9932 =.0068 Because p –value =.0068 <  =.05, we reject H 0.

23 Slide MA4704Gerry Golding n p –Value Approach p -value  p -value  0 0 z  = 1.645 z  = 1.645  =.05 z z z = 2.47 z = 2.47 One-Tailed Tests About a Population Mean:  Known Sampling distribution of Sampling distribution of

24 Slide MA4704Gerry Golding 5. Determine whether to reject H 0. We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes. Because 2.47 > 1.645, we reject H 0. Critical Value Approach Critical Value Approach One-Tailed Tests About a Population Mean:  Known For  =.05, z.05 = 1.645 4. Determine the critical value and rejection rule. Reject H 0 if z > 1.645

25 Slide MA4704Gerry Golding p -Value Approach to Two-Tailed Hypothesis Testing The rejection rule: The rejection rule: Reject H 0 if the p -value < . Reject H 0 if the p -value < . Compute the p -value using the following three steps: Compute the p -value using the following three steps: 3. Double the tail area obtained in step 2 to obtain the p –value. the p –value. 2. If z is in the upper tail ( z > 0), find the area under the standard normal curve to the right of z. the standard normal curve to the right of z. If z is in the lower tail ( z < 0), find the area under If z is in the lower tail ( z < 0), find the area under the standard normal curve to the left of z. the standard normal curve to the left of z. 1. Compute the value of the test statistic z.

26 Slide MA4704Gerry Golding Critical Value Approach to Two-Tailed Hypothesis Testing The critical values will occur in both the lower and The critical values will occur in both the lower and upper tails of the standard normal curve. upper tails of the standard normal curve. n The rejection rule is: Reject H 0 if z z  /2. Reject H 0 if z z  /2. Use the standard normal probability distribution Use the standard normal probability distribution table to find z  /2 (the z -value with an area of  /2 in table to find z  /2 (the z -value with an area of  /2 in the upper tail of the distribution). the upper tail of the distribution).

27 Slide MA4704Gerry Golding Example: Glow Toothpaste Two-Tailed Test About a Population Mean:  Known Two-Tailed Test About a Population Mean:  Known oz. Glow Quality assurance procedures call for Quality assurance procedures call for the continuation of the filling process if the sample results are consistent with the assumption that the mean filling weight for the population of toothpaste tubes is 6 oz.; otherwise the process will be adjusted. The production line for Glow toothpaste The production line for Glow toothpaste is designed to fill tubes with a mean weight of 6 oz. Periodically, a sample of 30 tubes will be selected in order to check the filling process.

28 Slide MA4704Gerry Golding Example: Glow Toothpaste Two-Tailed Test About a Population Mean:  Known Two-Tailed Test About a Population Mean:  Known oz. Glow Perform a hypothesis test, at the.03 Perform a hypothesis test, at the.03 level of significance, to help determine whether the filling process should continue operating or be stopped and corrected. Assume that a sample of 30 toothpaste Assume that a sample of 30 toothpaste tubes provides a sample mean of 6.1 oz. The population standard deviation is believed to be 0.2 oz.

29 Slide MA4704Gerry Golding 1. Determine the hypotheses. 2. Specify the level of significance. 3. Compute the value of the test statistic.  =.03 p –Value and Critical Value Approaches p –Value and Critical Value Approaches Glow H 0 :  H a : Two-Tailed Tests About a Population Mean:  Known

30 Slide MA4704Gerry Golding Glow Two-Tailed Tests About a Population Mean:  Known 5. Determine whether to reject H 0. p –Value Approach p –Value Approach 4. Compute the p –value. For z = 2.74, cumulative probability =.9969 p –value = 2(1 .9969) =.0062 Because p –value =.0062 <  =.03, we reject H 0. We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

31 Slide MA4704Gerry Golding Glow Two-Tailed Tests About a Population Mean:  Known  /2 =.015  /2 =.015 0 0 z  /2 = 2.17 z z  /2 =.015  /2 =.015 p -Value Approach p -Value Approach -z  /2 = -2.17 z = 2.74 z = -2.74 1/2 p -value =.0031 1/2 p -value =.0031 1/2 p -value =.0031 1/2 p -value =.0031

32 Slide MA4704Gerry Golding Critical Value Approach Critical Value Approach Glow Two-Tailed Tests About a Population Mean:  Known 5. Determine whether to reject H 0. We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz. Because 2.47 > 2.17, we reject H 0. For  /2 =.03/2 =.015, z.015 = 2.17 4. Determine the critical value and rejection rule. Reject H 0 if z 2.17

33 Slide MA4704Gerry Golding  /2 =.015 0 0 2.17 Reject H 0 Do Not Reject H 0 z z Reject H 0 -2.17 Glow Critical Value Approach Critical Value Approach Sampling distribution of Sampling distribution of Two-Tailed Tests About a Population Mean:  Known  /2 =.015

34 Slide MA4704Gerry Golding Confidence Interval Approach to Two-Tailed Tests About a Population Mean Select a simple random sample from the population Select a simple random sample from the population and use the value of the sample mean to develop and use the value of the sample mean to develop the confidence interval for the population mean . the confidence interval for the population mean . (Confidence intervals are covered in Chapter 8.) (Confidence intervals are covered in Chapter 8.) If the confidence interval contains the hypothesized If the confidence interval contains the hypothesized value  0, do not reject H 0. Otherwise, reject H 0. value  0, do not reject H 0. Otherwise, reject H 0.

35 Slide MA4704Gerry Golding The 97% confidence interval for  is The 97% confidence interval for  is Confidence Interval Approach to Two-Tailed Tests About a Population Mean Glow Because the hypothesized value for the Because the hypothesized value for the population mean,  0 = 6, is not in this interval, the hypothesis-testing conclusion is that the null hypothesis, H 0 :  = 6, can be rejected. or 6.02076 to 6.17924

36 Slide MA4704Gerry Golding n Test Statistic Tests About a Population Mean:  Unknown This test statistic has a t distribution with n - 1 degrees of freedom. with n - 1 degrees of freedom.

37 Slide MA4704Gerry Golding n Rejection Rule: p -Value Approach H 0 :   Reject H 0 if t > t  Reject H 0 if t < - t  Reject H 0 if t t  H 0 :   H 0 :   Tests About a Population Mean:  Unknown n Rejection Rule: Critical Value Approach Reject H 0 if p –value < 

38 Slide MA4704Gerry Golding p -Values and the t Distribution The format of the t distribution table provided in most The format of the t distribution table provided in most statistics textbooks does not have sufficient detail statistics textbooks does not have sufficient detail to determine the exact p -value for a hypothesis test. to determine the exact p -value for a hypothesis test. However, we can still use the t distribution table to However, we can still use the t distribution table to identify a range for the p -value. identify a range for the p -value. An advantage of computer software packages is that An advantage of computer software packages is that the computer output will provide the p -value for the the computer output will provide the p -value for the t distribution. t distribution.

39 Slide MA4704Gerry Golding A State Highway Patrol periodically samples A State Highway Patrol periodically samples vehicle speeds at various locations on a particular roadway. The sample of vehicle speeds is used to test the hypothesis Example: Highway Patrol One-Tailed Test About a Population Mean:  Unknown One-Tailed Test About a Population Mean:  Unknown The locations where H 0 is rejected are deemed The locations where H 0 is rejected are deemed the best locations for radar traps. H 0 :  < 65

40 Slide MA4704Gerry Golding Example: Highway Patrol One-Tailed Test About a Population Mean:  Unknown One-Tailed Test About a Population Mean:  Unknown At Location F, a sample of 64 vehicles shows a At Location F, a sample of 64 vehicles shows a mean speed of 66.2 mph with a standard deviation of 4.2 mph. Use  =.05 to test the hypothesis.

41 Slide MA4704Gerry Golding One-Tailed Test About a Population Mean:  Unknown 1. Determine the hypotheses. 2. Specify the level of significance. 3. Compute the value of the test statistic.  =.05 p –Value and Critical Value Approaches p –Value and Critical Value Approaches H 0 :  < 65 H a :  > 65

42 Slide MA4704Gerry Golding One-Tailed Test About a Population Mean:  Unknown p –Value Approach p –Value Approach 5. Determine whether to reject H 0. 4. Compute the p –value. For t = 2.286, the p –value must be less than.025 (for t = 1.998) and greater than.01 (for t = 2.387)..01 < p –value <.025 Because p –value <  =.05, we reject H 0. We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. of vehicles at Location F is greater than 65 mph.

43 Slide MA4704Gerry Golding Critical Value Approach Critical Value Approach 5. Determine whether to reject H 0. We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap. Because 2.286 > 1.669, we reject H 0. One-Tailed Test About a Population Mean:  Unknown For  =.05 and d.f. = 64 – 1 = 63, t.05 = 1.669 4. Determine the critical value and rejection rule. Reject H 0 if t > 1.669

44 Slide MA4704Gerry Golding  0 0 t  = 1.669 t  = 1.669 Reject H 0 Do Not Reject H 0 t One-Tailed Test About a Population Mean:  Unknown

45 Slide MA4704Gerry Golding End of Chapter 9, Part A

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